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Difference between revisions of "2007 iTest Problems/Problem 24"

m (Created page with "== Problem == Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube. Find the remainder w...")
 
(Solution to Problem 24)
 
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== Problem ==
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==Problem==
  
 
Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube.
 
Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube.
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\text{(G) }6 \quad
 
\text{(G) }6 \quad
 
\text{(H) }7 \quad
 
\text{(H) }7 \quad
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\text{(I) } 8\quad \\ </math>
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\text{(I) } 8\quad</math>
  
 
<math>\text{(J) }9 \quad
 
<math>\text{(J) }9 \quad
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\text{(O) }14 \quad
 
\text{(O) }14 \quad
 
\text{(P) }15 \quad
 
\text{(P) }15 \quad
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\text{(Q) }16 \quad \\ </math>
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\text{(Q) }16 \quad</math>
  
 
<math>\text{(R) }17 \quad
 
<math>\text{(R) }17 \quad
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\text{(X) }23 </math>
 
\text{(X) }23 </math>
  
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== Solution ==
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==Solution==
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The prime factorization of <math>2008</math> is <math>2^3 \cdot 251</math>, and the [[prime factorization]] of <math>2007</math> is <math>3^2 \cdot 223</math>.  Since <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube, the exponents in the prime factorization of <math>2008N</math> are even, and the exponents in the prime factorization of <math>2007N</math> are a multiple of three.
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 +
To make the exponents of <math>2008N</math> even, the exponent of <math>2</math> in <math>N</math> is at least <math>1</math>, but since there are no powers of <math>2</math> in <math>2007</math>, the exponent of <math>2</math> in <math>N</math> is at least <math>3</math>.  Similarly, in <math>N</math>, the exponent of <math>251</math> is at least <math>3</math>, the exponent of <math>3</math> is at least <math>4</math>, and the exponent of <math>223</math> is at least <math>2</math>.
 +
 
 +
Thus, <math>N</math>, the minimum positive integer that satisfies the criteria, equals <math>2^3 \cdot 3^4 \cdot 251^3 \cdot 223^2</math>.  Using [[modular arithmetic]] to find the remainder,
 +
<cmath>N \equiv 8 \cdot 6 \cdot 1^3 \cdot (-2)^2 \pmod{25}</cmath>
 +
<cmath>N \equiv 192 \pmod{25}</cmath>
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<cmath>N \equiv 17 \pmod{25}</cmath>
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The remainder when <math>N</math> is divided by <math>25</math> is <math>\boxed{\textbf{(R) } 17}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=23|num-a=25}}
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 19:49, 30 June 2018

Problem

Let $N$ be the smallest positive integer such that $2008N$ is a perfect square and $2007N$ is a perfect cube. Find the remainder when $N$ is divided by $25$.

$\text{(A) }0 \quad \text{(B) }1 \quad \text{(C) }2 \quad \text{(D) }3 \quad \text{(E) }4 \quad \text{(F) }5 \quad \text{(G) }6 \quad \text{(H) }7 \quad \text{(I) } 8\quad$

$\text{(J) }9 \quad \text{(K) }10 \quad \text{(L) }11 \quad \text{(M) }12 \quad \text{(N) }13 \quad \text{(O) }14 \quad \text{(P) }15 \quad \text{(Q) }16 \quad$

$\text{(R) }17 \quad \text{(S) }18 \quad \text{(T) }19 \quad \text{(U) }20 \quad \text{(V) }21 \quad \text{(W) }22 \quad \text{(X) }23$

Solution

The prime factorization of $2008$ is $2^3 \cdot 251$, and the prime factorization of $2007$ is $3^2 \cdot 223$. Since $2008N$ is a perfect square and $2007N$ is a perfect cube, the exponents in the prime factorization of $2008N$ are even, and the exponents in the prime factorization of $2007N$ are a multiple of three.

To make the exponents of $2008N$ even, the exponent of $2$ in $N$ is at least $1$, but since there are no powers of $2$ in $2007$, the exponent of $2$ in $N$ is at least $3$. Similarly, in $N$, the exponent of $251$ is at least $3$, the exponent of $3$ is at least $4$, and the exponent of $223$ is at least $2$.

Thus, $N$, the minimum positive integer that satisfies the criteria, equals $2^3 \cdot 3^4 \cdot 251^3 \cdot 223^2$. Using modular arithmetic to find the remainder, \[N \equiv 8 \cdot 6 \cdot 1^3 \cdot (-2)^2 \pmod{25}\] \[N \equiv 192 \pmod{25}\] \[N \equiv 17 \pmod{25}\] The remainder when $N$ is divided by $25$ is $\boxed{\textbf{(R) } 17}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 23 		Followed by:
Problem 25
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